Saved Bookmarks
This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If $ \cos \theta+\sin \theta=\sqrt{2} \cos \theta, $ show that$ \cos \theta-\sin \theta=\sqrt{2} \sin \theta $ |
|
Answer» cosA+sinA=√2cosAsquaring both the sides =>(cosA+sinA)²=2cos²A=>cos²A+sin²A+2sinAcosA=2cos²A=>cos²A-2cos²A+2sinAcosA= -sin²A=> -cos²A+2sinAcosA= -sin²A=> cos²A-2sinAcosA=sin²Aadding sin²A on both the sides => cos²A+sin²A-2sinAcosA=2sin²A=> (cosA-sinA)²=2sin²A=> cosA-sinA=√2sinA |
|
| 2. |
\frac { \sin 2 \theta } { 1 + \cos 2 \theta } \cdot \frac { \cos \theta } { 1 + \cos \theta } = \tan \theta / 2 |
| Answer» | |
| 3. |
In the sequence of numbers 2, 8, 26, 62, 122, 212, x,...., the term x is |
| Answer» | |
| 4. |
| संख्या क्रम 2, 8, 26, 62, 122, 212, ऋ, ....... में पद T |
| Answer» | |
| 5. |
कु 212 + 5#8)\‘ |
|
Answer» 8y^4/4u+6y^3/4u+4u^2/4u+4u/4u=2y^4/u+3/2y^3/u+u+1 |
|
| 6. |
(क) 212 (-3) |
|
Answer» -7 is the correct answer. -7 is the right answer of this question, can you agree with my answer, comment plz😍😍😍😍 -7 is the right ans in this case If we multiplying or dividing with any negative number the answer will come negative so the answer of above questions is -7 -7 is the write answer in this case itna bhi ni pta kya tujhevaise ans -7 hai 👍👍👍 -7 is the correct answer of the given question 21÷(-3)=answer -7 thank you 21 divided -3= correct answer is -7 -7 is correct answer -7 is the correct answer of the given question -7 is the correct answer of the given question -7 is a right answer this question |
|
| 7. |
If $ \cos \theta+\sin \theta=\sqrt{2} \cos \theta, $ show that $ \cos \theta-\sin \theta=\sqrt{2} \sin \theta $ |
| Answer» | |
| 8. |
At midnight the water at a particular beach is at high tide. At the same time a gauge at theend of a pier reads 10 feet. Low tide is reached at 6 AM when the gauge reads 4ft.Determine the amplitude. |
|
Answer» Amplitude = 10-4= 6feet |
|
| 9. |
V28 -212 cm |
|
Answer» √28² - 21² √ (28 - 21) ( 28 + 21) √ 7 × 49 7 √7 |
|
| 10. |
<olनै A4 |
|
Answer» =7.1/2+2/5-4.1/3=35+4/10-4/35=117-40/30=77/30 =2.56666666 15/2+2/5-13/3LCM of 2,5,3=30225/30 + 12/30 -130/30237/30-130/30107/30 7/2+2/5-12/334/10-12/392-120/30-28/30=0.9333 answer 15/2+2/5-13/3225/30+12/30-130/30237/30-130/30107/30 15/2 + 2/5 -- 13/3225+12-130 / 30107/30is correct answer |
|
| 11. |
рдирд┐74 a4+ |
|
Answer» Rationalising the denominatorhence7-4√3*(7-4√3)/(7+4√3)(7-4√3)=49+48-56√3/49-48=97-56√3/1hence a =97 and b=-56 |
|
| 12. |
tan(A)/(-cot(A) %2B 1) %2B cot(A)/(-tan(A) %2B 1)=1 %2B 1/(sin(A)*cos(A)) |
| Answer» | |
| 13. |
Find the area of a parallelogram ABCD and hence the altitude on BC |
| Answer» | |
| 14. |
/2+3+ 2128. Show that |
|
Answer» By doing rationalization(√2+1)(√2+1)/2-12+1+2√2/13+2√2 |
|
| 15. |
8. Show thatV2 +1-3+212 |
|
Answer» hit like if you find it useful |
|
| 16. |
18. In fig, ABCD and ABEF are parallelograms. The area of the Parallelogram ABCD is 90 sq cm. Find(a) ar (ABEF)(b) ar (ABD)(c) ar (BEF)OR |
|
Answer» Area of given ABCD parallelogram is 90 cm. (i) area of rectangle ABEF = area of parallelogram ABCD because , one common side is AB of parallelogram and rectangle and hight of both parallelogram is same as Breadth of rectangle .hence, area of rectangle ABEF = 90 cm² (ii) area of ABD = half of area of parallelogram { see attachment it is clear that , diagonal is divided by two equal of parallelogram, here BD is diagonal , that's why are(ABD) = ar(ABCD)/2 } hence, ar(ABD) = 90cm²/2 = 45 cm² (iii) ar(BEF) = half of ar(ABEF) = 45 cm² [ similarly diagonal of rectangle divide its two equal part ] |
|
| 17. |
(cosec*theta - cos(theta))^2=(-cos(theta) %2B 1)/(cos(theta) %2B 1) |
|
Answer» spelling is mistake please correct |
|
| 18. |
1/(cos(A) %2B 1) %2B 1/(-cos(A) %2B 1)=2*(A*cosec^2) |
| Answer» | |
| 19. |
sqrt((sec(theta) %2B 1)/(sec(theta) - 1)) %2B sqrt((sec(theta) - 1)/(sec(theta) %2B 1))=2*(cosec*theta) |
|
Answer» 2 is the correct answer 2cosec is the correct answer of this question |
|
| 20. |
1, 2 and 3, respecueryWrite the denominator of the rational number 257/5000 in the form 2m x 5n, where m, n are non-negativeintegers. Hence, write its decimal expansion, without actual division. |
|
Answer» 257/5000=257/2*2*2*5*5*5*5m=3,n=4decimal expansion=0.0514 257/5000=257/2*2*2*5*5*5*5m=3,n=4decimal expansion=0.0514 |
|
| 21. |
tan(theta)/(sec(theta) - 1)=(tan(theta) %2B sec(theta) %2B 1)/(tan(theta) %2B sec(theta) - 1) |
|
Answer» So we can write it assec -1/ tan = tan + sec-1/tan +sec+1 tonow As tan/sec-1= 1+sin a/cosa |
|
| 22. |
Smallest cell in size observed in bacteria is |
|
Answer» So your answer is Amoeba |
|
| 23. |
Write these numbers in order of size.Start with the smallest number.0.920.9010.99 |
|
Answer» in order of size 0.901,0.92,0.99 |
|
| 24. |
Write these numbers in order of size.Start with the smallest number.0.920.9010.990.0990.909 |
|
Answer» 0.92 0.901 0.99 0.099 0.909= The order of size with the smallest to the largest is 0.099 0.901 0.909 0.92 0.99 0.909 is the correct answer order number 0.92 to 0.909 is the correct answer 0.099 to 0.99 is the correct answer |
|
| 25. |
19. ABCD is a paraliclogram. E is a point on BA such that BE -2EA and Fis a point on DC wuch2FC. Prove that AECE is a parallelogram whose arca is one-third of the area otparallelogram ABCD. |
| Answer» | |
| 26. |
ez, the cap used by the Turks, is shaped like theA ffrustum of a cone (see Fig. 13.24). If its radius on theopen side is 10 cm, radius at the upper base is 4 cmand its slant height is 15 cm, find the area of materialused for making it. |
| Answer» | |
| 27. |
without actual division whether following number have terminating decimal expansion or non-terminating but repeating expansion a. 425 upon 625 |
|
Answer» 625 = 5^4 * 2^0Since 625 can be expressed in the form (2^m * 5^n), the fraction is terminating. |
|
| 28. |
to The number of sides of two regular polygons are as 5 : 4 and the differgnce bargles is 9 Find the number of sides of the polygons11. A rail road curve is to be laidout |
| Answer» | |
| 29. |
% यदि \tan \theta=\frac{3}{4} \frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+3 \cos \theta} |
| Answer» | |
| 30. |
\tan \theta=\frac{4}{3} \frac{2 \sin \theta-3 \cos \theta}{2 \sin \theta+3 \cos \theta}=-\frac{1}{17} |
|
Answer» ok |
|
| 31. |
\frac { ( \operatorname { cos } 2 \theta + i \operatorname { sin } 2 \theta ) ^ { 3 } ( \operatorname { cos } 4 \theta - i \operatorname { sin } 4 \theta ) ^ { 3 } } { ( \operatorname { cos } \theta + i \operatorname { sin } \theta ) ^ { 3 } } |
| Answer» | |
| 32. |
(-2*sin(theta)^3 %2B sin(theta))/(2*cos(theta)^3 - cos(theta))=tan(theta) |
| Answer» | |
| 33. |
6.The price of a motorcycle was 60,000 last year. It isincreased by 12% this year due to increase in price ofraw materials. Find the new price of the motorcycle. |
|
Answer» 67,200 is the right answer 67, 200 is the right answer the prize of motorcycle was = 60000prize was increased by 12% hence,12%of 60000 + 6000012/100×60000+600007200+6000067200 answer |
|
| 34. |
In Fig e t,ys, then prove that Aiis ล line.Fig. 6.16In Fig. 6.17, POQ is a line. Ray OR is perpendicularto line PQ. OS is another ray lying between raysOP and OR. Prove that5.2o d y is produced |
| Answer» | |
| 35. |
3 \sin \theta + 5 \cos \theta = 5 , \text { prove that } 5 \sin \theta - 3 \cos \theta = \pm 3 |
| Answer» | |
| 36. |
nd the fraction(AI CISE13. Rewrite the following as a quadratic equation in x and then solve for x(AI CBS2x + 3 |
| Answer» | |
| 37. |
A fez, the cap used by the Turks, is shaped like thefrustum of a cone (see Fig. 13.24). If its radius on theopen side is 10 cm, radius at the upper base is 4 cmand its slant height is 15 cm, find the area of materialused for making it.d made un of |
| Answer» | |
| 38. |
A fez, the cap used by the Turks, is shaped like thefrustum of a cone (see Fig. 13.24). If its radius on theopen side is 10 cm, radius at the upper base is 4 cmand its slant height is 15 cm, find the area of materialused for making it.3.d made up of a met |
|
Answer» love you |
|
| 39. |
-41m +11 mn + 18h का गुणनखण्ड क्या होगा ? |
|
Answer» (m+2n) (m+2n) is the right answer |
|
| 40. |
The sum of n terms of two AP's are in the ratio (5n+ 4): (9n + 16), find the ratio of their 18hterms. |
|
Answer» thanks |
|
| 41. |
9.The number of sides of two regular polygons are in the ratio of 3 :2 and their interior angles are in theratio of 5: 3. Find the number of their sides. |
|
Answer» Let sides of 1st polygon=3A,sides of 2nd polygon=2A, but now according to question, we have [(3A-2)×180°/3A]/[(2A-2)×180°/2A] = 5/3, [(3A-2)×2A]/[(2A-2)×3A] = 5/3, (6A-4)/(6A-6) =5/3, 18A-12=30A-30, 30A-18A=-12+30, 12A=18, A=3/2 |
|
| 42. |
of two regular polygons having che sides1,then find the9. Ο e angle of a pentagon is 1400 If the remaining angles are in the ratio 4 : 2 : 3and the largest anglesuoo whichthe |
|
Answer» write the polygons having 7 sides write the name of the polygons having 7 side |
|
| 43. |
10. Two regular polygons are such that the ratio between their number of sidesis 1:3 and the ratio of the measures of their interior angles is 3:4. Find theHOTSnumber of sides of each polygon |
|
Answer» Let the sides be N₁& N₂ N₁:N₂ = 1:3 (N₁-2): (N₂-2) =3:4 SOLVE N₁ = 5 & N₂ = 6 |
|
| 44. |
(-tan(x) %2B 1)/(tan(x) %2B 1) |
|
Answer» tan(π/4 -x)is right answer |
|
| 45. |
\frac { \operatorname { sec } \theta - \operatorname { tan } \theta } { \operatorname { sec } \theta + \operatorname { can } \theta } = 1 - 2 \operatorname { sec } \theta \operatorname { tan } \theta + 2 \operatorname { tan } ^ { 2 } \theta |
| Answer» | |
| 46. |
16. For a given A. P. T30-T25=50. Then, for that AP d is____A, 25 B.5 C.20 D.1017. The tenth term of the A. P. √3, 3√3, 5√3, 7√3. .. is.......A. 12 B. 11 3 C. 19 3 D. 103 |
| Answer» | |
| 47. |
(\sin \theta+\sec \theta)^{2}+(\cos \theta+\csc \theta)^{2}=(1+\sec \theta \csc \theta)^{2} |
| Answer» | |
| 48. |
() 12 cm,98area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find thelensth of the altitude lcorresponding torthe shortest side.d sbsm ai alldmu nA os61125 mef this field is sold'at thethe |
|
Answer» Semiperimeter (s) = (42 + 34 + 20 ) / 2So it is = 96 / 2 = 48 cm. Using herons formula = (√s )(√s-a )(√s-b) (√s-c)By substituting s and a , b , c i.e the given sides.We have√48 ×√6 ×√14 ×√28So √(4 × 6 × 2) ×√6 ×√(7 × 2) ×√(7 × 4)Simplifying the numbers we have area = 4 × 6 × 2 × 7 = 336 cm²Hence area is 336 cm² Now height corresponding to longest side implies that base is 42 cm and area remains same So area of triangle is 1/ 2 b × h336 = 1/ 2 × 42 × HHENCE H = 16 cm So height is 16 cm . |
|
| 49. |
(5*sin(theta) - 2*sec(theta)^3 %2B 2*cos(theta))/(5*sin(theta) %2B 2*sec(theta)^3 - 2*cos(theta)) |
| Answer» | |
| 50. |
sqrt((-tan(x) %2B 1)/(tan(x) %2B 1)) |
| Answer» | |